The rate constant for a first order reaction is given by the following equation: ln k = 33.24 - 2.0 x 10^4 K / T The activation energy for the reaction is given by ____________ kJ mol^(-1). (In nearest integer) (Given : R = 8.3 J K^(-1) mol^(-1))

ln k=ln A-frac E_ A RT Given: ln k=33.24-frac2.0 x 10^4 T Therefore on comparing frac E_ A R=2.0 x 10^4 \ Therefore E_ A=2.0 x 10^4 x R \ => E_ A=2.0 x 10^4 x 8.3 J \ => E_ A=16.6 x 10^4 J=166 kJ

The rate constant for a first order reaction is given b... | null (Chemistry)

\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}

Given:

\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}

\begin{aligned} &\therefore \text { on comparing } \frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{R}}=2.0 \times 10^4 \\\\ &\therefore \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times \mathrm{R} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ} \end{aligned}

Explanation

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